Solve using substitution: $\frac {dy}{dx}-\frac{4y}{x}=2x^5e^{\frac {y}{x^4}}$ I need to solve this by using a substitution. So far I've used $u=\frac {y}{x^4}$ and $u=e^{\frac {y}{x^4}}$ but the first doesn't let me equate $Mu$ to $Nx$ (method of integrating factor) and the second gets immensely complex when trying to take all these derivatives and this is only problem #5 in my homework so I'm sure that I'm doing something wrong (the more difficult problems generally begin after #40). Does anyone know something better to substitute and a general strategy for finding a substitution term? 
 A: You perfectly can do it using $u=\frac {y}{x^4}$ which means $y=x^4u$, $y'=4x^3u+x^4u'$. Now, replace in the original equation $$y'-\frac{4y}{x}=2x^5e^{\frac {y}{x^4}}$$ and you get $$x^4u'=2x^5 e^u$$ Divide by $x^4$, put all the $u$' on the same side and get $$u'e^{-u}=2x$$ which is separable and that you can rewrite as $$-\Big(e^{-u}\Big)'=\Big(x^2\Big)'$$ Integrate and get $$-e^{-u}=x^2+C$$ $$e^{-u}=C-x^2$$ Take logarithms of both sides $$-u=\log(C-x^2)$$ and finally $$u=-\log(C-x^2)$$ $$y=x^4u=-x^4\log(C-x^2)$$

If you want to use instead $v=e^{\frac {y}{x^4}}$ which means $y=x^4\log(v)$, $y'=\frac{x^4 v'}{v}+4 x^3 \log (v)$ then the original equation write $$\frac{x^4 v'}{v}=2x^5 v$$ that is to say $$\frac {v'}{v^2}=2x$$ or $$-\Big(\frac {1}{v}\Big)'=\Big(x^2\Big)'$$ $$\frac {1}{v}=-x^2+C$$ $$v=\frac 1{-x^2+C}$$ $$y=x^4\log(v)=-x^4\log(C-x^2)$$
A: Note that
$$\frac{d}{dx}\left(e^{-y/x^4}\right)=-e^{-y/x^4}\left(\frac{1}{x^4}\frac{dy}{dx}-\frac{4y}{x^5}\right)=\frac{-e^{-y/x^4}}{x^4}\left(\frac{dy}{dx}-\frac{4y}{x}\right)$$
Thus, we have 
$$\frac{dy}{dx}-\frac{4y}{x}=2x^4e^{y/x^4}\implies \frac{-e^{-y/x^4}}{x^4}\left(\frac{dy}{dx}-\frac{4y}{x}\right)=-2x\implies \frac{d}{dx}\left(e^{-y/x^4}\right)=-2x$$ 
whereupon solving for $y$ gives 
$$y=-x^4\log(C-x^2)$$
where $C$ is an integration constant.
